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A body of mass $6 \mathrm{~kg}$ is moving with a uniform velocity 4 $\mathrm{ms}^{-1}$. Its velocity changes to $6 \mathrm{~ms}^{-1}$ when a force of $12 \mathrm{~N}$ acts on it. Then its displacement is
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Verified Answer
The correct answer is:
$5 \mathrm{~m}$
(b) Mass of body $=6 \mathrm{~kg}$
Initial velocity, $\mathrm{u}=4 \mathrm{~m} / \mathrm{s}$
Final velocity, $v=6 \mathrm{~m} / \mathrm{s}$.
Force, $\mathrm{F}=12 \mathrm{~N}$
Using Newtones IInd law, F = ma
$$
\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2
$$
Using $\mathrm{Eq}^{\mathrm{n}}$ of Motion,
$$
\begin{aligned}
& v^2-u^2=2 a s ; s=\frac{v^2-u^2}{2 a} \\
& \Rightarrow s=\frac{(6)^2-(4)^2}{2 \times 2}=\frac{36-16}{4}=\frac{20}{4} \\
& \Rightarrow s=5
\end{aligned}
$$
Hence, displacement is $5 \mathrm{~m}$.
Initial velocity, $\mathrm{u}=4 \mathrm{~m} / \mathrm{s}$
Final velocity, $v=6 \mathrm{~m} / \mathrm{s}$.
Force, $\mathrm{F}=12 \mathrm{~N}$
Using Newtones IInd law, F = ma
$$
\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2
$$
Using $\mathrm{Eq}^{\mathrm{n}}$ of Motion,
$$
\begin{aligned}
& v^2-u^2=2 a s ; s=\frac{v^2-u^2}{2 a} \\
& \Rightarrow s=\frac{(6)^2-(4)^2}{2 \times 2}=\frac{36-16}{4}=\frac{20}{4} \\
& \Rightarrow s=5
\end{aligned}
$$
Hence, displacement is $5 \mathrm{~m}$.
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